Thursday, May 23, 2013

Problem 879: Square, Midpoint, Diagonal, Ratio 3:1, Angle Measure

Geometry Problem. Post your solution in the comments box below.
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the complete problem 879.

Online Geometry Problem 879: Square, Midpoint, Diagonal, Ratio 3:1, Angle Measure

Posted by Antonio Gutierrez at 8:06 AM
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7 comments:

  1. AnonymousMay 23, 2013 at 12:18 PM

    point F is 1/4 of the diagonal AC, drop a vertical from E and a horizontal from F gives us a base of 1 and a height of 3 for the triangle giving us an atan of 1/3 for half of the angle E. From here we find the tanE = 36.8698976..........

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    Replies
    1. AjitMay 25, 2013 at 9:18 PM

      Draw EG//AB meeting AC in G, let /_FEG=α & /_GED=β, x = α + β . Further, tan(α)=1/3 and tan(β)=1/2 and so tan(x)=tan(α+β)=[tan(α)+tan(β)]/[1-tan(α)*tan(β)]=[1/3 + 1/2] = [1-1/6]=1. Hence, x=45°

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  2. UnknownMay 23, 2013 at 2:36 PM

    EC=X
    DC=2X
    build BD
    AC=sqrt(5)x
    AF=y
    FC=3y
    4y=sqrt(5)x
    y=sqrt(5)x/4=AF
    BD cut AC at Q
    QF=sqrt(5)x/2-AF=sqrt(5)x/2-sqrt(5)x/4=sqrt(5)x/4
    FQ=AF
    DQ=sqrt(5)x/2
    DC/DQ=EC/FQ=2x/0.5sqrt(5)x=x/0.25sqrt(5)x=
    =4/sqrt(5)
    FQD~ECD
    DFQ=DEC
    F point on circle of ECD
    [can to prove it by "Reductio ad absurdum"]
    FCD=FED=45

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  3. Michael TsourakakisMay 23, 2013 at 3:55 PM

    http://img829.imageshack.us/img829/3494/22778697.png

    Angle EKF=angle KED=90+45=135 degrees.If,AB=a ,is, FK=AK/2=a.sqrt(2)/4,EK=a/2 ,KD=a.sqrt(2)
    FK/EK=(sqrt(2))/2 (1) ,EK/KD=(sqrt(2))/2 (2)
    From (1),(2) , FK/EK= EK/KD so, the triangles EKF,EKD is similar ,therefore, angle FEK= angle EDK=ω, angle KED= angle KFE=y. But ω+y=45 ,so, angle x=45 degrees

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  4. Peter TranMay 23, 2013 at 8:26 PM

    http://img208.imageshack.us/img208/6976/problem897.png
    Draw GFH //BA
    Since AF/AC=1/4 so AH=BG=GE=1/4.AD
    Triangle DHF congruence to FGE (case SAS)
    So FD=FE and angle(GFE)+ angle (HFD)= 90 => angle (EFA)=90
    Triangle EFD is a right isosceles triangle => angle (FED)=45

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  5. Sumith PeirisSeptember 9, 2015 at 1:59 AM

    From congruent triangles we see that FB = FE = FD so F is the centre of the circumcircle of Tr. BED. So < EFD = 2 < EBD = 90.
    Hence CDFE is cyclic and it follows that x = < FCD = 45.

    Sumith Peiris
    Moratuwa
    Sri Lanka

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  6. APOSTOLIS MANOLOUDISAugust 24, 2016 at 3:28 AM

    Problem 879
    If the BD intersects AC at O then CO=2AF=2FO or AF=FO.But OF/OD=1/2=EC/CD so triangle ECD is similar with triangle FOD. Then <CDE=<ODF but <CDE+<EDB=45 or
    <ODB+<EDO=45 or <EDF=45=<ECF so ECDF is cyclic. Therefore <EFD=90 and <FED=45.
    APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE

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